Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-4y &= 3 \\ -5x+2y &= 9\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-5x = -2y+9$ Divide both sides by $-5$ to isolate $x$ $x = {\dfrac{2}{5}y - \dfrac{9}{5}}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{2}{5}y - \dfrac{9}{5}}) - 4y = 3$ $-\dfrac{16}{5}y + \dfrac{72}{5} - 4y = 3$ Simplify by combining terms, then solve for $y$ $-\dfrac{36}{5}y + \dfrac{72}{5} = 3$ $-\dfrac{36}{5}y = -\dfrac{57}{5}$ $y = \dfrac{19}{12}$ Substitute $\dfrac{19}{12}$ for $y$ in the top equation. $-8x-4( \dfrac{19}{12}) = 3$ $-8x-\dfrac{19}{3} = 3$ $-8x = \dfrac{28}{3}$ $x = -\dfrac{7}{6}$ The solution is $\enspace x = -\dfrac{7}{6}, \enspace y = \dfrac{19}{12}$.